Activation Polarization

The Tafel equation (13-38), which relates the activation polarization component of the overvoltage to the current /, may be accounted for in terms of the following analysis. Consider a general electrode reaction

where, as illustrated in Fig. 13-13, the oxidized state must pass through some high-energy intermediate I in the process of being reduced. We write an equation analogous to Eq. (8-65) for the rate of the forward process:

(13-50)

e~ + Ο (oxidized state) = R (reduced state),

(O), (13-51)

where JG?* is the standard free energy change to go from Ο to I. It will be known

SPECIAL T O P I C S , S E C T I O N 2 531 Intermediate I

Reduced state R F I G . 13-13. Mechanism for activation polarization.

in the next chapter as the free energy of activation. Similarly, the rate of the back reaction is

*

^{b =}

fir

^ex

e(^4]

^(R)

-

⁽¹³

-

⁵²⁾

The presence of a potential difference φ between the electrode and the solution contributes to J G ? * and we write this last as the sum of a chemical component and an electrical one:

AG^ot = AG°%^hem - *&φ, (13-53)

which allows that only some of this potential difference is effective; this fraction is called the transfer coefficient oc. The current part due to the forward reaction is then

'-Κ^ΤΜ Μ Ή-)) <°>· <"-*>

where current is in faradays per square centimeter per second. At equilibrium there will be equal and opposite currents in the two directions, or if° = /b° = i°, where i° is known as the exchange current, and φ has the value φ° and corresponds to Then

Combination of Eqs. (13-54) and (13-55) gives

h = j° exp ^ l i o > = r exp ^ , (13-56)

where η is the overvoltage. Equation (13-56) may be written in the form ln, = ln/° + ^ - ,

which rearranges to the Tafel equation.

In a stricter analysis, an equation analogous to Eq. (13-54) is written for ή>,

N e g a t i v e 0 — P o s i t i v e

V

F I G . 13-14. A general current versus voltage diagram.

with ι = i^f — i^b; the effect is to alter the coefficient of η in Eq. (13-56) to give

As the equation suggests, the situation is symmetric; η may be positive or negative, so that the electrode reaction is driven either forward or backward. Figure 13-14 shows the general behavior of Eq. (13-57).

This treatment, although somewhat sketchy, is designed to indicate how the detailed study of overvoltage effects can lead to information about the intrinsic rate of the electrode reaction, through i°, and the energy of the reaction barrier.

As an example, although the mechanism for the reduction of H⁺ ion at a metal electrode is still not fully elucidated, the evidence suggests that the rate-determining step is probably the reaction of H³0 + with the metal surface to give adsorbed hydrogen atoms and water:

β. Concentration Polarization

The preceding analysis was based on the assumption that the rate-limiting step of the electrode process was some activated chemical reaction at the electrode surface. Another process must become rate-controlling at sufficiently high current densities, namely the rate of diffusion of reactant to and of product away from the electrode surface. If the reaction is one of the deposition of a metal from solution, only the rate of diffusion of the metal ion to the electrode is to be considered.

(13-57)

H³0 + + Μ + e - = M - H + H^zO .

SPECIAL TOPICS, SECTION 2 5 3 3

Recalling Eq. (2-65) and the discussion of Section 10-7B, we can write

dx (13-58)

where / is now in moles per square centimeter per second. If there is an excess of inert electrolyte in the solution so that the ion being reduced carries little of the current, the potential term in Eq. (12-121) will not be important and Of will be a constant, equal to the ion diffusion coefficient as given by Eq. (12-119). Further, a reasonable approximation to the physical situation is that the stirring conditions leave a thin stagnant layer of thickness δ within which C varies linearly from C, the bulk solution concentration, to C^s, the concentration at the electrode surface.

We expect the linear variation of C with χ since in a steady-state condition / is constant, and since S is constant, so must be dCjdx. Equation (13-58) then becomes

where i& is the diffusion current density in faradays per square centimeter per second and ζ is the valence number of the metal ion being discharged. Under the condi­

tions of the preceding subsection it is assumed that S/8 is large enough that C^s = C, and the observed current is determined by the rate of the chemical reduction reaction at the electrode, z^a . That is, under steady-state conditions there is no accumulation of material in the diffusion zone and i& = i&. As the over-voltage is increased, i^& increases by Eq. (13-57), and C^s drops enough for i^d to match the increase in i^a. The maximum possible diffusion rate i'd(max) is reached when C^s has dropped to zero. Further increase in the overvoltage cannot increase the current any further, and the plot of i versus V must level off at id(max), or at

Thus the effect of concentration polarization is to give a maximum current which is proportional to the concentration of the ion being discharged.

If a mixture of potentially reducible ions is present, again with a supporting electrolyte, that is, an excess of inert electrolyte, present, then as the applied reducing potential is increased the first metal ion begins to deposit, reaches its /max, and the second metal ion begins to deposit at some higher potential, and reaches its /^{m a x}, so that the plot of i versus V is as shown in Fig. 13-15. The characteristic deposition potential for each ion could be identified as in Fig. 13-9 but when one is operating in the diffusion-controlled region C^s and hence S^rew is changing with V so that one obtains a sinusoidal rather than a linear i versus V plot. Consequently, it is much more accurate to pick the potential at the half-way point of the step, or the half-wave potential K^{1 / 2}, as the characteristic one.

Since the overvoltage η is given by

/d = zJ = — zS(C* - C)

δ (13-59)

*d(max) — z®C

(13-60)

V rev — (13-61)

F I G . 1 3 - 1 5 . Current versus voltage in the region of concentration polarization.

where ohmic polarization is neglected, as are activity coefficient terms, we can solve for C^s/C. Also, combination of Eqs. (13-59) and (13-60) gives

The result is

id = id (max) y ^ r ) .

Id

id (max) = 1 — exp RT

(13-62)

(13-63) Equation (13-63) gives a symmetric curve such as shown in Fig. 13-15, for which

V^1/2 occurs at the inflection point.

C. Polarography

The preceding analysis suggests that we could analyze for metal ions in solution by obtaining experimental / versus V curves, each limiting diffusion current giving the concentration of a particular ion, the ion being identified by its V^1/2. This procedure is indeed used. To maintain δ more constant than is possible by stirring

F I G . 1 3 - 1 6 . Schematic diagram of a polarographic cell.

SPECIAL TOPICS, SECTION 2 535 the solution, a common practice is to use a rotating electrode. There is still a problem with accumulation of reduction products, and a very ingenious alternative procedure was devised by J. Heyrovsky in 1922.

The basic experimental features of the Heyrovsky polarograph are shown in Fig. 13-16. The cathode is a mercury drop that is steadily growing at the tip of a capillary immersed in the electrolyte solution, and the anode is a large pool of mercury. By having a tiny drop as the anode surface, concentration polarization effects can be made to develop at relatively small currents; the large area of the anode pool of mercury essentially eliminates concentration polarization at its surface. As each drop grows and falls the anode surface is steadily kept fresh and reproducible, and with a supporting electrolyte the same general analysis applies as for a stationary electrode.

A typical polarogram is shown in Fig. 13-17. The oscillations are a result of the successive appearance of new drops at intervals of a few seconds; currents may be only microamperes in order of magnitude. Each step is called a polarographic wave and is characterized by its half-wave potential K^{1 / 2} and diffusion current i'd(max) . The detailed algebraic analysis is complicated by the situation of an expanding cathode surface, and an equation derived by D. Ilkovic in 1938 will be given without the derivation:

id(m») = 7 0 . 8 2 z w ( f O ( ^ ) ( i C ) , (13-64) where ζ is the valence number of the ion, w is the flow of the mercury in grams

per second, / is the drop time in seconds, and C is the concentration in moles per liter. Equations have also been derived for the half-wave potential, but this remains essentially an empirically determined quantity for each species.

Polarography is capable of measuring ion concentrations as low as 10~⁴ Μ and is a rapid as well as a sensitive analytical tool. The physical chemist uses it to study the chemistry of reduction (or of oxidation) processes. One can determine from the diffusion current whether the reduction occurs as a one- or as a

two-V (relative to standard calomel electrode) F I G . 13-17. A typical polarogram.

electron step. If complexing ions are added to the solution, both the reduction potentials and the formation constants of complex ions can be found. With a commutator to interrupt the applied voltage, chemical rate processes can be followed.

A modern variant of polarography uses a fixed (not a dropping mercury) electrode and current is recorded as the voltage is changed at a fixed rate. In cyclic voltammetry the voltage sweep is first in one direction and then in the other. Sweep rates of up to 100 V s e c '¹ may be used. An idealized voltammogram is shown in Fig. 13-18. On the forward sweep (upper curve) reduction products are formed and on the backward sweep (lower curve) these are oxidized. About the same informa­

tion is obtained as in ordinary polarography but, in addition, much can be learned about the nature and the reaction kinetics of the reduction product.

250 /Liamps d i v^{- 1}

ic

0.125 V d i v¹ _|_ . J. 1 1

( + )

fa

1 1

o ( - )

v = 0.1 V s e c - i

FIG. 13-18. Typical cyclic voltammogram for a reversible redox couple. [From D. T. Sawyer and J. L. Roberts, Jr., "Experimental Electrochemistry for Chemists" Wiley {Interscience), New

York, 1974.]

G E N E R A L R E F E R E N C E S

ADAMSON, A . W . (1976). " T h e Physical Chemistry o f Surfaces," 3rd ed. Wiley (Interscience), N e w York.

DANIELS, F., MATHEWS, J. H., WILLIAMS, J. W . , BENDER, P., AND ALBERTY, R. A . (1956). "Experi­

mental Physical Chemistry," 5th ed. McGraw-Hill, N e w York.

DOUGLAS, Β . E., AND MCDANIEL, D . H. (1965). "Concepts and Models of Inorganic Chemistry."

Ginn (Blaisdell), Boston, Massachusetts.

KORTUM, G. (1965). "Treatise o n Electrochemistry." Elsevier, Amsterdam.

EXERCISES 537 LATIMER, W . M . (1952). " T h e Oxidation States o f the E l e m e n t s a n d Their Potentials i n A q u e o u s

Solution," 2nd ed. Prentice-Hall, Englewood Cliffs, N e w Jersey.

LEWIS, G. N . , AND RANDALL, M. (1961). "Thermodynamics," 2nd ed. (revised by K. S. Pitzer and L . Brewer). McGraw-Hill, N e w York.

C I T E D R E F E R E N C E S

BATES, R. G. (1954). "Electrometric pH Determinations." Wiley, N e w York.

DOLE, M. (1941). "The Glass Electrode." Wiley, N e w York.

HARNED, H. S., AND EHLERS, R. W. (1932). / . Amer. Chem. Soc. 5 4 , 1350.

LATIMER, W. M. (1952). "The Oxidation States o f the Elements and Their Potentials in Aqueous Solution," 2nd ed. Prentice-Hall, Englewood Cliffs, N e w Jersey.

E X E R C I S E S

Activity coefficient effects are t o b e neglected i n Exercises a n d P r o b l e m s unless specifically n o t e d otherwise. A s s u m e 2 5 ° C unless otherwise specified. T a k e as exact numbers given t o o n e significant figure.

13-1 T h e e m f o f the cell P b / P b C l2( s ) / K C l ( m ) / H g2C l2( s ) / H g is 0.5357 V at 25°C a n d increases with temperature b y 1.45 x 1 0 "4 V K ~1. Write t h e electrode a n d overall cell reactions and calculate AW J 5 ° , and qTev .

Ans. P b + H g2C l2 = P b C l2 4- 2Hg(/), J G ° = - 1 0 3 . 4 kJ, J 5 ° = 2 8 . 0 J Κ "1, ΔΗ0 = - 9 5 . 1 kJ, qT„ = 8.34 kJ.

13-2 T h e e m f for the cell Cd/0.1 m C d ( N Oa)2, 0.01 m A g N Os/ A g is - 0 . 4 4 6 V at 25°C.

Calculate <f° for this cell.

Ans. - 0 . 3 5 7 V.

13-3 Calculate <f298 for the cell A g / A g u » / 0 . 1 m H I / H2( 1 atm)/Pt.

Ans. 0.038 V .

13-4 Calculate <?° and Κ at 25°C for the reaction Tl + Ag+ = T1+ + A g .

Ans. <?° = 1.135 V, Κ = 1.56 χ 1 01 9.

13-5 Calculate Κ for the reaction £ C d + T1+ = £ C d2+ + Tl and the ratio ( C d2 +) / ( T l+) if excess C d is added to a solution which is 0.1 m in T l+.

Ans. Κ = 13.4 Μ "1 / 2; 2.76.

13-6 Calculate <f°⁹⁸ for the half-cell reaction Tl = T l3 + -f 3e~.

Ans. - 0 . 7 2 1 V.

13-7 Obtain < ^9 8 for the half-cell M g + C2OJ" = M g ( C204) + 2e~ given that KBp = 9.0 χ 10~ 6

for M g ( C204) .

Ans. 2.49 V.

13-8 What fraction o f A g ( C N )2~ is dissociated into A g+ in 1 χ ΙΟ"3 Μ C N ~ if <?°9B = 0.289 V for the half-cell A g + 2 C N ~ = A g ( C N )2- + e"? (Neglect the hydrolysis o f C N ~ . )

Ans. 4 . 0 χ 1 0 -1 8( ! ) .

13-9 Calculate &2 9 8 for the following concentration cells: (a) N a ( H g , j cN a = 0.1)/NaCl(m = 0.1)/Na(Hg, j tN a = 0.01); (b) Pt/H2(/> = 0.1 atm)/HCl(m = 0Λ)/Η2(Ρ = 0.01 atm)/Pt;

(c) Ag/AgBr(s)/HBr(m = 0.1)/H2(1 atm)/Pt—Pt/H2(l atm)/HBr(m = 0.01)/AgBr(i)/Ag.

Ans. (a) 0.0592 V , (b) 0.0296 V , (c) 0.1183 V .

13-10 Calculate &3 5 0 for the cell in Exercise 13-9(b).

Ans. 0.0348 V.

13-11 Calculate <f2 9 8 for the cell Ag/AgCl(j)/0.1 m HC1/H2(1 atm)/Pt, using activity coefficient data from Chapter 12.

Ans. - 0 . 3 5 2 V.

13-12 A pH meter gives a reading of 200 m V when measuring a solution whose hydrogen ion activity is considered to be 1 0- 4; what is the hydrogen ion activity of a solution for which the meter reads 1 0 0 m V ? (Assume 25°C.)

(a) Calculate the solubility product for CuCl in water at 25°C.

(b) What is the concentration of cuprous ion at the anode of this cell?

13-3 Given that £ ° = 0.152 for A g + I" = A g l + e~ at 25°C and is - 0 . 8 0 0 for A g = Ag+ + e~, calculate the solubility product for A g l .

13-4 The voltage of the cell Ag/Ag2S04(.s)/saturated solution of A g2S 04 and H g2S 04/ H g2S 04/ H g is 0.140 V at 25°C and its temperature coefficient is 0.00015 V °Cr\

(a) Give the cell reaction.

(b) Calculate the free energy change for the cell reaction.

(c) Calculate the enthalpy change for the cell reaction.

(d) Calculate the entropy change for the cell reaction.

(e) D o e s the cell absorb or emit heat as the cell reaction occurs ? Calculate the number of calories per mole of cell reaction.

(f) O n e mole each of H g , A g2S 04( j ) , H g2S 04( j ) , and s o m e saturated solution of the t w o

PROBLEMS 539 w h o s e emf is - 0 . 7 6 7 V at 25°C:

(a) Write the electrode and net cell reactions.

(b) Calculate the equilibrium constant for A g + + 2 C N- = A g ( C N )2~ (neglect activity coefficient effects). T h e d o u b l e dashed diagonal denotes a salt bridge w h i c h (it is h o p e d ) m a k e s the j u n c t i o n potential negligible.

13-8 At 25°C, € for the cell P b / P b S O4( 5) / H2S O4( 0 . 0 0 2 0 0 w ) / H2( l atm)/Pt is 0.125 V. Calculate the solubility product of PbSO* at 25°C.

13-9 Calculate the percentage of mercury in the mercuric state in a solution of mercuric nitrate that is in equilibrium with liquid mercury.

13-10 Calculate the solubility product of ferrous hydroxide from the cell F e ( j ) / F e ( O H)2( 5 ) / Ba(OH)2(0.05 m ) / H g O( 5) / H g . ^2 9 8 = 0.973 V.

13-11 M o l e c u l a r weights are given i n k g i n the SI s y s t e m ; thus, the molecular weight o f Oa

is 0.032 kg. I n g o i n g f r o m the c g s t o the SI system, state whether the numerical value o f e a c h o f the f o l l o w i n g is c h a n g e d , a n d if it is, calculate the n e w value, (a) A v o g a d r o ' s number, (b) Faraday's number, (c) < f ° for the cell P t / H2( l a t m ) / H C l / A g C l / A g . (d) AG0

for the cell reaction in (c). (e) £ at 25°C for the cell P t / H2( l atm)/HCl(0.001 m ) / A g C l / A g . W h a t w o u l d y o u r answers be h a d the molecular weight o f 02 been defined as 32 k g m o l e ~1

(and m h a d been k g molecular weights per 1000 k g o f s o l v e n t ) ?

13-12 It is desired to separate C d2 + from P b2 + by electrodeposition from a solution which is 0.1 m in each ion and at pH 2. Calculate the sequence in which Cd and Pb metals and H2

are produced and the concentrations of the various species in solution when a new stage of electrolysis occurs. Include overvoltage effects in considering H2 evolution.

13-13 R. Ogg found * = - 0 . 0 2 9 V at 25°C for the cell

What is the formula for mercurous i o n ? S h o w h o w your conclusion follows from this information.

T h e actual dissociation constant for Hg2"*" = 2Hg+ is not k n o w n , but suppose that £ for this cell is found to be —0.059 V when the higher and lower mercurous nitrate nor­

malities are 2 χ 10"* and 3 χ 10~β, respectively. Calculate KdiBB for H g2* from this data.

(In both parts of the problem neglect any junction potential.)

13-14 Calculate the emf of the following cells (neglect activity coefficient effects):

(c) Ag/AgBr/HBr(0.01 w ) / H2( 0 . 5 atm)/Pt.

Neglect junction potentials.

13-15 If copper and silver are negligibly soluble in each other as solids, approximately what must be the concentration o f A g+ in a solution in which C u2 + is at unit activity in order that the t w o metals may be plated out together in electrolysis ? A t what potential will this occur, o n the assumption that oxygen is formed at the anode at an overvoltage of 0.50 V at 1 atm pressure from a solution of pH 5 ? A s s u m e 25°C.

(b) (a)

13-16 At 25°C, / for the cell P t / H²( l a t m ) / H C l ( m ) / A g C l / A g is as follows (Harned and Ehlers, 1932):

m 0.01002 0.01010 0.01031 0.04986 0.05005 0.09642

<f(V) 0.46376 0.46331 0.46228 0.38582 0.38568 0.35393 m 0.09834 0.2030

* ( V ) 0.35316 0.31774

Calculate £ ° for the cell by the extrapolation method, and then the activity coefficients for HCl at these concentrations and plot them against Vrh; compare the limiting slope with the value from theory.

13-17 At 25°C, £ for the cell

A g / A g C l / N a C l ( m ) / N a H g ( a m a l g a m ) — N a H g (amalgam)/NaCl(0.100 m)/AgCl/Ag is as follows:

m 0.200 0.500 1.000 2.000 3.000 4.000

^ 2 9 8 (V) 0.03252 0.07584 0.10955 0.14627 0.17070 0.19036 The mean activity coefficient of 0.100 m N a C l is 0.773. Calculate the mean activity coefficients of N a C l in solutions of these concentrations and plot the values against the square root of m.

13-18 At 25°C, * for the cell P b / P b S 0⁴( s ) / H²S 0⁴( m ) / H²( l atm)/Pt changes with the molality of the sulfuric acid as follows:

m 0.00100 0.00200 0.00500 0.0100 0.0200

* 2 9 β Ο 0 0.1017 0.1248 0.1533 0.1732 0.1922 By the extrapolation method, obtain ^° for this cell. F r o m this value calculate the thermo­

dynamic solubility product for P b S 0⁴.

13-19 At 25°C, / for the cell P t / H²( l atm)/ m Ba(OH)²/Ba(amalgam)/0.1200 m B a ( O H )²/ H² (1 atm)/Pt is as follows:

1000 m 2.45 7.93 15.09 49.70 199.60

* ( V ) - 0 . 1 2 4 5 - 0 . 0 8 4 0 - 0 . 0 6 3 0 - 0 . 0 2 5 4 + 0 . 0 1 4 4 (a) Write the cell reaction (each electrode separately and then the net reaction).

(b) By means of the appropriate extrapolation procedure, calculate the mean activity coefficient for 0.1000 m B a ( O H )².

13-20 Suppose that the atomic weight of carbon is defined as 12 kg m o l e^{- 1}. Explain whether this redefinition would affect the values o f <f and # ° for the cell Pb/PbCl²(s)/KCl(10 g liter"¹)/

H g²C l²( j ) / H g at 25°C.

13-21 The p H of a solution may be found with the use of a hydrogen electrode and a calomel half-cell. Suppose that the pH of a certain buffer solution is measured at a high altitude such that the ambient pressure is 600 Torr but that the observer neglects this aspect and calculates the pH assuming the pressure of the hydrogen bubbling past the electrode to be 1 atm. If he reports a pH of 5.50, what is the correct pH of the solution?

13-22 Calculate the free energy of formation of AgCl(j) using the solubility product for AgCl and data from Table 13-3.

13-23 Calculate AG⁰ for the reaction H^aO ( / ) = H+ + O H ~ at 25°C given that the free energy of formation of Tiquid water is —56.69 kcal m o l e^{- 1} and using Table 13-3.

13-24 A thallous (T1+) solution is titrated with eerie ( C e^{4 +}) solution. The reaction is T1+ + 2Ce⁴+ = Tl³+ + 2 C e^{8 +}.

In document ELECTROCHEMICAL CELLS (Pldal 32-43)